Answer:
[tex]T_o = 141.81 ^0C[/tex]
Explanation:
Given that;
Thermal diffusivity [tex]\alpha = 5 \times 10 ^{-6} m^2/s[/tex]
Thermal conductivity [tex]k = 20 \ W/m.K[/tex]
Heat transfer coefficient h = ( we are to assume the imposed surface temperature ) = 20 W/m².K
Initial temperature = 150 ° C = (150+273) K = 423 K
Then coolant temperature with which the casting is exposed to = 20° C = (20+273)K = 293 K
Time = 40 seconds
Length = 20mm = 0.02 m
The objective is to determine the temperature at the surface at a depth of 20 mm after 40 seconds.
[tex]Bi = \dfrac{hL}{k}[/tex]
[tex]Bi = \dfrac{20*0.02}{20}[/tex]
Bi == 0.02
[tex]\tau = \dfrac{\alpha t}{L^2}[/tex]
[tex]\tau= \dfrac{5*10^{-6 }* 40}{0.020^2}[/tex]
[tex]\tau = 0.5[/tex]
For a wall at 0.2 Bi
[tex]A_1 = 1.0311[/tex]
[tex]\lambda _1 = 0.4328[/tex]
Therefore;
[tex]\dfrac{T_o - T_{\infty}}{T_i - T_{\infty}}= A_1 e ^{-( \lambda_1^2 \ \tau)[/tex]
[tex]\dfrac{T_o - 293 }{423 - 293}= 1.0311 \times e ^{-( 0.438^2 \times 0.5 )[/tex]
[tex]\dfrac{T_o - 293 }{423 - 293}= 1.0311 \times e ^{-( 0.0959 )[/tex]
[tex]\dfrac{T_o - 293 }{130}= 1.0311 \times 0.9085[/tex]
[tex]\dfrac{T_o - 293 }{130}= 0.937[/tex]
[tex]T_o - 293= 0.937 \times 130[/tex]
[tex]T_o - 293= 121.81[/tex]
[tex]T_o = 121.81+ 293[/tex]
[tex]T_o = 414.81 \ K[/tex]
[tex]T_o = (414.81 - 273)^0C[/tex]
[tex]T_o = 141.81 ^0C[/tex]
